Answer :

Given:




T2; T3& T4


Since T2 , T3 & T4 are in AP


Then; 2(T3) = T2 + T4


i.e.







(n-1)(n-2)-6(n-1)+6=0


n2-3n+2-6n+6+6=0


n2-9n+14=0


(n-2)(n-7)=0


n= 2,7


n=2 rejected for term 3rd


So n=7

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